Solution
Step 1
Write an expression for the mean
[tex]\bar{x}=\text{ }\frac{\sum ^{\square}_{\text{ of terms}}}{\text{mumber of terms}}[/tex][tex]\begin{gathered} \bar{x}=\frac{2+2+1+\text{ 4 +3+3+3+3}+4+1}{10} \\ \bar{x}=\text{ }\frac{26}{10} \\ \bar{x}=2.6 \end{gathered}[/tex]Step 2
Find the standard deviation
[tex]SD=\sqrt[]{\frac{\sum |x-\bar{x}|^2}{n}}[/tex][tex]SD=\sqrt[]{\frac{|2-2.6|^2+|2-2.6|^2+|1-2.6|^2+|4-2.6|^2+|3-2.6|^2+|3-2.6|^2+|3-2.6|^2+|3-2.6|^2+|4-2.6|^2+|1-2.6|^2}{10}}[/tex][tex]\begin{gathered} SD=\sqrt[]{\frac{0.36+0.36+2.56+1.96+0.16+0.16+0.16+0.16+1.96+2.56}{10}} \\ =\sqrt[]{\frac{10.4}{10}}\text{ =}\sqrt[]{1.04} \\ =1.02 \end{gathered}[/tex]Find the Z (confidence interval level) value for 90%
[tex]Z_{90}=1.645[/tex]Find the confidence interval
[tex]\begin{gathered} CI=\bar{x}\pm_{}_{}z\times\frac{s}{\sqrt[]{n}} \\ \bar{x}\Rightarrow\operatorname{mean} \\ s\Rightarrow\text{standard deviation} \\ n\Rightarrow\text{sample size} \\ z\Rightarrow\text{confidence level} \end{gathered}[/tex][tex]\begin{gathered} CI=2.6\pm1.645\times\frac{1.02}{\sqrt[]{10}} \\ =2.6\pm1.645\times0.32 \\ =2.6\pm0.53 \\ \text{Thus,} \\ CI=2.6+0.53\text{ 0r 2.6-0.53} \\ CI=3.13\text{ or 2.07} \end{gathered}[/tex]Hence, a 90% confidence interval estimate of the mean lies between 2 and 3
Therefore, either Michelangelo or Donatello